1
Question
Evaluate: ∫sinx√36−cos2xdx
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Solution
We have, ∫sinx√36−cos2xdx
Let cosx=t⇒−sinxdx=dt
Putting sinxdx=−dt and cosx=t.
=−∫dt√36−t2
=−∫dt√(6)2−t2
=−sin−1(t6)+c
=−sin−1(cosx6)+c ....[∵t=cosx]
Let cosx=t⇒−sinxdx=dt
Putting sinxdx=−dt and cosx=t.
=−∫dt√36−t2
=−∫dt√(6)2−t2
=−sin−1(t6)+c
=−sin−1(cosx6)+c ....[∵t=cosx]
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